Question

Equal circles with centres O and O’ touch each other at X. OO’ produced to meet a circle with centre O’, at A. AC is tangent to the circle whose centre is O. O’ D is perpendicular to AC. Find the value of .

Answer 1

Answer:

DO'/CO=1/3

This can be done by proving ADO' and ACO similar..

A is common

O'D and CO are perpendiculars respectively..

By AA similarlity

Since both circles are equal AO=3r

AO' is r

r/3r=O'D/CO..

hence proved..

please mark it brainliest

Answer 2

we know that ∠ADO' = 90° ( since O'D is perpendicular to AC)

∠ACO= 90° ( OC(radius)perpendicular to AC(tangent))

In triangles ADO'and ACO ,

∠ADO' = ∠ACO ( each 90°)

∠DAO = ∠CAO (common)

by AA criterion ,triangles ADO' and ACO are similar to each other.

AO'/AO = DO'/CO ( corresponding sides of similar triangles )

AO = AO' + O'X + OX

= 3AO' (since AO'=O'X=OX because radii of the two circles are equal )

AO'/AO = AO'/3AO =1/3

DO'/CO=AO'/AO = 1/3

DO'/CO =1/3.